I have to answer three questions about limits of Riemann sums, and I can't remember how to figure it out. I know it involves sequences with sigma but I don't know where to start now.
the first one I have is written with the integral from 2 to 5 (x^2 - 1)dx
can someone please get me started or show the steps so I can do the rest of my homework?
As the function is continuous everywhere it is so in the given interval and thus it is Riemann integrable there. Now subdivideOriginally Posted by ipreferhistory
subintervals of equal length
, and choose to evaluate the function
at the left endpoints of each subinterval (we can do that since we know the functions is Riemann int. and thus we can freely choose the partition ,as long as it length parameter tends to zero when
, and the points at each subinterval at which the function's evaluated:
![\int\limits_2^5(x^2-1)dx=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0} ^{n-1}\left[\left(2+\frac{3k}{n}\right)^2-1\right]=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0}^{n-1}\left[4+\frac{12}{n}k+\frac{9}{n^2}k^2-1\right]](http://latex.codecogs.com/png.latex?\int\limits_2^5(x^2-1)dx=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0} ^{n-1}\left[\left(2+\frac{3k}{n}\right)^2-1\right]=\lim_{n\to\infty}\frac{3}{n}\sum\limits_{k=0}^{n-1}\left[4+\frac{12}{n}k+\frac{9}{n^2}k^2-1\right])
Tonio
The question reads:
Find a formula for right hand endpoint sum by dividing the interval into n equal subintervals. Then take the limit of these sums as n approaches infinity to calculate area under the curve [a,b]. f(x)=5x^2 over [1,2]
  |
LinkBack URL
About LinkBacks